where \(\mathbf {Q}\)  is the orthogonal eigenvector matrix, i.e. \(\mathbf {Q}^{=\mathbf {Q}^{-1}, and \mathbf {\Lambda } = \text{diag}(\lambda _1,\lambda _2,\dots ) contains the real eigenvalues. And the solution to Eq \ref {eq:5} can be solved using the standard eigenvalue equation:\begin{equation} \bf {FQ = Q\Lambda }, \quad \text{with} \quad \det (\mathbf {F}-\lambda \mathbf {I})=0\end{equation}}The Williamson’s theorem provides us with a symplectic variant of the results above. Let \) F R2n 2n\( be a symmetric and positive definite matrix, the Williamson’s theorem says that \) F\( can be diagonalized using symplectic matrices \cite {Nicacio2021Williamson, Gosson2006Symplectic}:\begin{equation} \bf S^{\mathsf {T}}FS = \hat{D} =\begin{bmatrix}\mathbf {D}& \\& \mathbf {D}\end{bmatrix} \end{equation}\) where \(\mathbf {D} = \text{diag}(d_1,d_2,\dots ,d_n)\)  is a diagonal matrix with positive entries (\(d_j\)  maybe zero if \(\mathbf {F}\)  is semidefinite). The \(d_j, \: j=1,2,\dots ,n\)  are said to be the symplectic eigenvalues of matrix \(\mathbf {F}\)  [1]} and are in general not equal to the eigenvalues given in Eq REF . The matrix \(\mathbf {S} = [u_1,\dots ,u_n,v_1,\dots ,v_n]\)  is a real symplectic matrix that satisfies the condition:
\( \bf S^{\mathsf {T}}JS = J, \quad \text{where} \quad \mathbf {J}=\begin{bmatrix}\mathbf {0_n} & \mathbf {I_n} \\-\mathbf {I_n} & \mathbf {0_n}\end{bmatrix} \) 
